Complete the statement.

hunter has a total of $9.60 in dimes and nickels. let n be the number of nickels and d be the number of dimes. hunter knows ther

Vladimir79 [104]

9514 1404 393

Answer:

(n, d) = (56, 68)

Hunter has 56 nickels and 68 dimes.

Step-by-step explanation:

The equations can be solved by using the second equation to substitute for d in the first equation.

10(n +12) +5n = 960

120 +15n = 960 . . . . . . . eliminate parentheses

15n = 840 . . . . . . . . . . . . subtract 120

n = 56 . . . . . . . . . . . . . . . divide by 15

d = 56 +12 = 68 . . . . . . . find d

The solution is n = 56, d = 68. This means Hunter has 56 nickels and 68 dimes.

_____

The meaning of the solution comes from the definitions of the variables.

8 0

2 years ago

Solve the compound inequality.

larisa [96]

Answer:

Step-by-step explanation:

9 - 4x > = 5 or 4(-1 + x) - 6 > = 2

-4x > = 5 - 9 -4 + 4x - 6 > = 2

-4x > = -4 4x - 10 > = 2

x < = -4/-4 4x > = 2 + 10

x < = 1 4x > = 12

x > = 12/4

x > = 3

answer is : B

7 0

3 years ago

What is the greatest common factor of 4k, 18k4, and 12?

Semenov [28]

The greatest common factor of 4k, 18k4 and 12 is
4

Finding the GCF of monomials requires inspection of the terms and looking for common factors that would be able to reduce the terms to whole numbers and variables with positive exponents.

5 0

2 years ago

Read 2 more answers

What is the value of 3-2

kicyunya [14]

Answer:

1

Step-by-step explanation:

brainliest pls if this helps

4 0

2 years ago

Find the mass of the lamina that occupies the region D = {(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1} with the density function ρ(x, y) = xye

Alona [7]

Answer:

The mass of the lamina is 1

Step-by-step explanation:

Let $\rho(x,y)$ be a continuous density function of a lamina in the plane region D,then the mass of the lamina is given by:

$m=\int\limits \int\limits_D \rho(x,y) \, dA$ .

From the question, the given density function is $\rho (x,y)=xye^{x+y}$ .

Again, the lamina occupies a rectangular region: D={(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1}.

The mass of the lamina can be found by evaluating the double integral:

$I=\int\limits^1_0\int\limits^1_0xye^{x+y}dydx$ .

Since D is a rectangular region, we can apply Fubini's Theorem to get:

$I=\int\limits^1_0(\int\limits^1_0xye^{x+y}dy)dx$ .

Let the inner integral be: $I_0=\int\limits^1_0xye^{x+y}dy$ , then

$I=\int\limits^1_0(I_0)dx$ .

The inner integral is evaluated using integration by parts.

Let $u=xy$ , the partial derivative of u wrt y is

$\implies du=xdy$

and

$dv=\int\limits e^{x+y} dy$ , integrating wrt y, we obtain

$v=\int\limits e^{x+y}$

Recall the integration by parts formula: $\int\limits udv=uv- \int\limits vdu$

This implies that:

$\int\limits xye^{x+y}dy=xye^{x+y}-\int\limits e^{x+y}\cdot xdy$

$\int\limits xye^{x+y}dy=xye^{x+y}-xe^{x+y}$

$I_0=\int\limits^1_0 xye^{x+y}dy$

We substitute the limits of integration and evaluate to get:

$I_0=xe^x$

This implies that:

$I=\int\limits^1_0(xe^x)dx$ .

Or

$I=\int\limits^1_0xe^xdx$ .

We again apply integration by parts formula to get:

$\int\limits xe^xdx=e^x(x-1)$ .

$I=\int\limits^1_0xe^xdx=e^1(1-1)-e^0(0-1)$ .

$I=\int\limits^1_0xe^xdx=0-1(0-1)$ .

$I=\int\limits^1_0xe^xdx=0-1(-1)=1$ .

No unit is given, therefore the mass of the lamina is 1.

3 0

3 years ago